Question

i need to find an equation of a parabola with the dimensions of the st louis arch. its 630ft high and 630ft wide.

Answer 1

Draw your parabola. Label the maximum value of y. Identify the y-intercept of the parabola. Determine the value of x at which y = 0.

Consider the equation y=ax^2 + bx + c OR y = c*x^2 + k.

Which one would be the easier to use here?

Answer 2

the first one

Answer 3

\(\bf recall\implies y={\color{red}{ a}}(x-h)^2+k\qquad vertex\ (h,k)\)

Answer 4

JDoe's illustration and his equation are both very helpful. Would you, @seesawn, please identify the vertex of the parabola you're trying to describe with an equation.

Referring to JDoe's equation, above, what is h? what is k?

Knowing the coordinates of the vertex, and the right x-intercept, how would you go about finding the value of the coefficient " a " in his equation?

Answer 5

what does a equal?

Answer 6

h=0 k=630

Answer 7

yes

Answer 8

so what's "a" ? well, we dunno
but we know from the points (-315, 0) and (315, 0) that when x = \(\pm 350\) y = 0

so we could just use either point to get "a" thus \(\bf y={\color{red}{ a}}(x-0)^2+630\qquad vertex\ (0,630)
\\ \quad \\
x = -315\qquad y=0\implies y={\color{red}{ a}}(x-0)^2+630\implies 0={\color{red}{ a}}(630-0)^2+630\) solve for "a"

Answer 9

ahemm rather when x = \(\pm 315\) y = 0 I meant =)

anyhow, so you'd use either point to solve for "a"

Answer 10

hmmm I even used the wrong haemm lemme fix all that,shoot, the typos

Answer 11

\(\bf y={\color{red}{ a}}(x-0)^2+630\qquad vertex\ (0,630)
\\ \quad \\
x = -315\qquad y=0\implies y={\color{red}{ a}}(x-0)^2+630\implies 0={\color{red}{ a}}(-315-0)^2+630\)

Answer 12

shoot... ahemm ok \(\bf y={\color{red}{ a}}(x-0)^2+630\qquad vertex\ (0,630)
\\ \quad \\
x = -315\qquad y=0\implies y={\color{red}{ a}}(x-0)^2+630
\\\quad \\\implies 0={\color{red}{ a}}(-315-0)^2+630\)

Answer 13

so using the point (-315, 0) and the vertex values, solve for "a" :)

Answer 14

would it be 157.5?
or did i do that wrong

Answer 15

well... ahemm.... sure?

Answer 16

if i got it right, the equation would be y=157.5(-315-0)^2+630

Answer 17

right?

Answer 18

hmmm well.. ahemm 157.5?

Answer 19

\(\bf y={\color{red}{ a}}(x-0)^2+630\qquad vertex\ (0,630)
\\ \quad \\
x = -315\qquad y=0\implies y={\color{red}{ a}}(x-0)^2+630
\\\quad \\\implies 0={\color{red}{ a}}(-315-0)^2+630\)
... what would you get for "a"?

Answer 20

idk, i subtracted the 630, then distributed the square, divided the answer from the parenthesis, and got 157.5

Answer 21

hmmm the 630 is outside the parenthesis, recall your PEMDAS :)

Answer 22

shiiiiizzzzzzzzz

Answer 23

so wouldnt that give me 99855?

Answer 24

yes, it would

Answer 25

well, actaully no

Answer 26

anyhow -> \(\bf x = -315\qquad y=0\implies y={\color{red}{ a}}(x-0)^2+630
\\\quad \\\implies 0={\color{red}{ a}}(-315-0)^2+630
\\ \quad \\
0=a(-315)^2+630\implies -630=a(-315)^2\implies \cfrac{-630}{(-315)^2}=a\)

Answer 27

-0.00634920634?

Answer 28

then you can simplify that, and that would be "a" :)

well, you may want to leave it as fraction, rather than decimal, either will work I gather

Answer 29

ok
thanks for the help!!!

Answer 30

\(\bf 0=a(-315)^2+630\implies -630=a(-315)^2\implies \cfrac{-630}{(-315)^2}=a
\\ \quad \\
-\cfrac{2}{315}=a\qquad thus
\\ \quad \\
y={\color{red}{ a}}(x-0)^2+630\quad vertex\ (0,630)\implies y={\color{red}{ -\cfrac{2}{315}}}(x-0)^2+630\)

Answer 31

yw